Difference between revisions of "Symmedians, Lemoine point"
(→Proportions) |
(→Parallel lines) |
||
(13 intermediate revisions by the same user not shown) | |||
Line 16: | Line 16: | ||
<cmath>\angle BDA = \angle ACB = \angle ACM \implies \triangle ABD \sim \triangle AMC \implies</cmath> | <cmath>\angle BDA = \angle ACB = \angle ACM \implies \triangle ABD \sim \triangle AMC \implies</cmath> | ||
<cmath>\frac {AM}{MC}= \frac {AB}{BD}.</cmath> | <cmath>\frac {AM}{MC}= \frac {AB}{BD}.</cmath> | ||
− | Similarly <math>\triangle | + | Similarly <math>\triangle ABM \sim \triangle ADC \implies \frac {AM}{MB}= \frac {AC}{CD}.</math> |
<cmath>BM = MC \implies \frac {BD}{CD} = \frac{AB}{AC}.</cmath> | <cmath>BM = MC \implies \frac {BD}{CD} = \frac{AB}{AC}.</cmath> | ||
Line 25: | Line 25: | ||
2. <math>\frac {BD}{CD} = \frac{AB}{AC}.</math> | 2. <math>\frac {BD}{CD} = \frac{AB}{AC}.</math> | ||
+ | |||
As point <math>D</math> moves along the fixed arc <math>BC</math> from <math>B</math> to <math>C</math>, the function <math>F(D) = \frac {BD}{CD}</math> monotonically increases from zero to infinity. This means that there is exactly one point at which the condition is satisfied. In this case, point <math>D</math> lies on the symmedian. | As point <math>D</math> moves along the fixed arc <math>BC</math> from <math>B</math> to <math>C</math>, the function <math>F(D) = \frac {BD}{CD}</math> monotonically increases from zero to infinity. This means that there is exactly one point at which the condition is satisfied. In this case, point <math>D</math> lies on the symmedian. | ||
Similarly for point <math>E.</math> | Similarly for point <math>E.</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | Let <math>AE</math> be the <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | Then <math>BE</math> is the <math>B-</math> symmedian of <math>\triangle ABD, CE</math> is the <math>C-</math> symmedian of <math>\triangle ACD, DE</math> is the <math>D-</math> symmedian of <math>\triangle BCD.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Symmedian and tangents== | ||
+ | [[File:Tangents and symmedian.png|220px|right]] | ||
+ | Let <math>\triangle ABC</math> and it’s circumcircle <math>\Omega</math> be given. | ||
+ | |||
+ | Tangents to <math>\Omega</math> at points <math>B</math> and <math>C</math> intersect at point <math>F.</math> | ||
+ | |||
+ | Prove that <math>AF</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>D = AF \cap \Omega \ne A.</math> WLOG, <math>\angle BAC < 180^\circ.</math> | ||
+ | <cmath>\triangle FDB \sim \triangle FBA \implies \frac {BD}{AB} = \frac{DF}{BF}.</cmath> | ||
+ | <cmath>\triangle FDC \sim \triangle FCA \implies \frac {CD}{AC} = \frac{DF}{CF}.</cmath> | ||
+ | <math>BF = CF \implies \frac {BD}{CD} = \frac{AB}{AC} \implies AD</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | [[File:Tangents to symmedian.png|220px|right]] | ||
+ | Let <math>\triangle ABC</math> and it’s circumcircle <math>\Omega</math> be given. | ||
+ | |||
+ | Let tangent to <math>\Omega</math> at points <math>A</math> intersect line <math>BC</math> at point <math>F.</math> | ||
+ | |||
+ | Let <math>FD</math> be the tangent to <math>\Omega</math> different from <math>FA.</math> | ||
+ | |||
+ | Then <math>AD</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Lemoine point properties== | ||
+ | [[File:L and G.png|390px|right]] | ||
+ | Let <math>\triangle ABC</math> be given. Let <math>L</math> be the Lemoine point of <math>\triangle ABC.</math> | ||
+ | |||
+ | <cmath>LD \perp BC, D \in BC, LE \perp AC, E \in AC, LF \perp AB, F \in AB.</cmath> | ||
+ | |||
+ | Prove that <math>\frac{LD}{BC} = \frac{LE}{AC} = \frac{LF}{AB}, L</math> is the centroid of <math>\triangle DEF.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>G</math> be the centroid of <math>\triangle ABC, GD' \perp BC, D' \in BC,</math> | ||
+ | <cmath>LE' \perp AC, E' \in AC, LF' \perp AB, F' \in AB.</cmath> | ||
+ | |||
+ | The double area of <math>\triangle AGC</math> is <math>2[AGC] = GE' \cdot AC = 2[BGC] = GD' \cdot BC \implies \frac {GD' }{GE' } = \frac {AC}{BC}.</math> | ||
+ | |||
+ | Point <math>L</math>is the isogonal conjugate of point <math>G</math> with respect to <math>\triangle ABC \implies \frac {LE}{LD} =\frac {GD' }{GE' } =\frac {AC}{BC}.</math> | ||
+ | |||
+ | Similarly, one can get <math>\frac {LE}{AC} = \frac {LD}{BC} = \frac {LF}{AB} = k.</math> | ||
+ | |||
+ | The double area of <math>\triangle DLE</math> is <math>2[DLE] = LD \cdot LE \sin \angle DLE = k BC \cdot k AC \cdot \sin \angle ACB = k^2 \cdot 2[ABC].</math> | ||
+ | |||
+ | Similarly, one can get <math>[DLE] = [DLF] = [DEF] = k^2 [ABC] \implies L</math> is the centroid of <math>\triangle DEF.</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | Vector sum <math>\vec {LE} + \vec {LD} + \vec {LF} = \vec 0.</math> | ||
+ | |||
+ | Each of these vectors is obtained from the triangle side vectors by rotating by <math>90^\circ</math> and multiplying by a constant <math>k^2,</math> | ||
+ | <cmath>\vec {AC} + \vec {CB} + \vec {BA} = \vec 0.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Parallel lines== | ||
+ | [[File:Symmedians perp and par.png|390px|right]] | ||
+ | Let <math>\triangle ABC</math> and it’s Lemoine point <math>L</math> be given. | ||
+ | |||
+ | Let <math>D</math> be an arbitrary point. Let <math>D'</math> be the foot from <math>D</math> to line <math>\overline{AC}</math>. | ||
+ | |||
+ | Denote <math>\ell</math> the line through <math>D</math> and parallel to <math>AC.</math> | ||
+ | |||
+ | Denote <math>\ell'</math> the line parallel to <math>AB</math> such that distance <math>EE' = DD' \cdot \frac {AB}{AC}</math> and points <math>E</math> and <math>D</math> are both in the exterior (interior) of <math>\triangle ABC.</math> | ||
+ | |||
+ | Prove that points <math>F = \ell \cap \ell', A,</math> and <math>L</math> are collinear. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>P(Q)</math> the foot from <math>L</math> to <math>\overline{AB}(\overline{AC})</math>. | ||
+ | |||
+ | <cmath>\frac {PL}{AB} = \frac {QL}{AC} \implies \frac {PL}{EE'} = \frac {QL}{DD'}.</cmath> | ||
+ | Denote <math>F = AL \cap \ell, F' = AL \cap \ell' \implies</math> | ||
+ | <cmath>\frac {FA}{AL} = \frac {DD'}{QL} = \frac {EE'}{PL} = \frac {F'A}{AL} \implies F = F'.</cmath> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | If squares <math>ABGF</math> and <math>ACDE</math> are constructed in the exterior of <math>\triangle ABC,</math> then <math>AO,</math> where <math>O</math> is the center of circle <math>\odot AEF,</math> is the symmedian in <math>\triangle ABC</math> through <math>A.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Radical axis== | ||
+ | [[File:Circles symmedians.png|390px|right]] | ||
+ | Circle <math>\omega</math> passes through points <math>A</math> and <math>B</math> and touches line <math>AC,</math> circle <math>\theta</math> passes through points <math>A</math> and <math>C</math> and touches line <math>AB.</math> Let <math>L</math> be the Lemoine point of <math>\triangle ABC.</math> | ||
+ | |||
+ | Prove that the radical axis of these circles contains the symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote centers of <math>\omega, \theta,</math> and <math>\odot ABC</math> throught <math>Q,P,</math> and <math>O,</math> respectively. | ||
+ | |||
+ | Denote <math>\ell</math> line throught <math>P</math> parallel to <math>AC, \ell'</math> line throught <math>Q</math> parallel to <math>AB.</math> | ||
+ | <math>\angle QAB = |90^\circ - \angle BAC| = \angle PAC \implies \triangle ABQ \sim \triangle ACP \implies</math> | ||
+ | |||
+ | The ratio of distance from <math>P</math> to <math>AC</math> to <math>AC</math> is equal to the ratio of distance from <math>Q</math> to <math>AB</math> to <math>AB \implies X = \ell \cap \ell' \in AL.</math> | ||
+ | <math>AP \perp AB \implies AP \perp \ell', QA \perp AC \implies QA \perp \ell \implies</math> | ||
+ | |||
+ | <math>A</math> is the orthocenter of <math>\triangle PQX \implies XA \perp QP \implies</math> | ||
+ | |||
+ | the radical axis of these circles contains the <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Common Lemoine point== | ||
+ | [[File:L to L.png|440px|right]] | ||
+ | [[File:2 46 Prasolov.png|440px|right]] | ||
+ | Let <math>\triangle ABC</math> be given, <math>\Omega = \odot ABC.</math> | ||
+ | |||
+ | Let <math>L</math> be the Lemoine point of <math>\triangle ABC.</math> | ||
+ | |||
+ | <math>A' = AL \cap \Omega \ne A, B' = BL \cap \Omega \ne B, C' = CL \cap \Omega \ne C.</math> | ||
+ | |||
+ | Prove that the point <math>L</math> is the Lemoine point of <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote point <math>D</math> so that <math>LD \perp BC, D \in BC.</math> | ||
+ | |||
+ | Similarly denote <math>E \in AC</math> and <math>F \in AB.</math> | ||
+ | <math>L</math> is the centroid of <math>\triangle DEF.</math> | ||
+ | |||
+ | <math>\triangle DEF \sim \triangle A'B'C'</math> (see Claim). | ||
+ | |||
+ | Let point <math>G</math> be the centroid of <math>\triangle A'B'C' \implies</math> | ||
+ | <cmath>\angle LDE = \angle GA'B'.</cmath> | ||
+ | <math>CDLE</math> is cyclic so <math>\angle LDE = \angle LCE = \angle LCA = \angle C'CA = \angle C'A'A = \angle C'A'L</math> | ||
+ | therefore <math>A'L</math> and <math>A'G</math> are isogonals with respect <math>\angle C'A'B'.</math> | ||
+ | |||
+ | Similarly <math>B'L</math> and <math>B'G</math> are isogonals with respect <math>\angle A'B'C' \implies</math> | ||
+ | |||
+ | <math>L</math> is the isogonal conjugate of a point <math>G</math> with respect to a triangle <math>\triangle A'B'C'</math> | ||
+ | |||
+ | so <math>L</math> is the Lemoine point of <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | <i><b>Claim</b></i> | ||
+ | |||
+ | Lines AP, BP and CP intersect the circumcircle of <math>\triangle ABC</math> at points <math>A', B',</math> and <math>C'.</math> | ||
+ | |||
+ | Points <math>D, E,</math> and <math>F</math> are taken on the lines <math>BC, CA,</math> and <math>AB</math> so that <math>\angle PDB = \angle PFA = \angle PEC</math> (see diagram). | ||
+ | |||
+ | Prove that <math>\triangle A'B'C' \sim \triangle DEF.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>\angle PFA = \angle PDB \implies PDBF</math> is cyclic so <math>\angle PDF = \angle PBF = \angle ABB' = \angle AA'B'.</math> | ||
+ | |||
+ | Similarly, <math>\angle PDE = \angle AA'C' \implies</math> | ||
+ | <math>\angle FDE = \angle PDF + \angle PDE = \angle AA'B' + \angle AA'C' = \angle B'A'C'.</math> | ||
+ | |||
+ | Similarly, <math>\angle DEF = \angle A'B'C'. \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Lemoine point extreme properties== | ||
+ | Lemoine point <math>L</math> minimizes the sum of the squares of the distances to the sides of the triangle (among all points internal to <math>\triangle ABC.)</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let us denote the desired point by <math>X.</math> Let us imagine that point <math>X</math> is connected to springs of equal stiffness attached to the sides at points <math>D, E,</math> and <math>F</math> and contacts sliding along them without friction. The segments modeling the springs will be perpendicular to the corresponding side. The energy of each spring is proportional to the square of its length. The minimum energy of the system corresponds to the minimum of the sum of the squares of the lengths of these segments, that is, the sum of the squares of the distances from <math>X</math> to the sides. | ||
+ | |||
+ | It is known that the minimum spring energy corresponds to the equilibrium position. The condition of equilibrium at a point <math>X</math> is the equality to zero of the vector sum of forces applied from the springs to the point <math>X.</math> The force developed by each spring is proportional to its length, that is, the equilibrium condition is that the sum of the vectors <math>\vec {XE} + \vec {XD} + \vec {XF} = \vec 0.</math> It is clear that the point <math>L</math> corresponds to this condition. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Lemoine point and perpendicularity== | ||
+ | [[File:Symmedians perp.png|430px|right]] | ||
+ | Let <math>\triangle ABC</math> be given. Let <math>L</math> be the Lemoine point of <math>\triangle ABC,</math> | ||
+ | <math>LD \perp BC, D \in BC, LE \perp AC, E \in AC, LF \perp AB, F \in AB.</math> | ||
+ | |||
+ | <math>M</math> is the midpoint <math>BC.</math> | ||
+ | |||
+ | Prove that <math>FE \perp AM.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>AL</math> is isogonal conjugated <math>AM</math> with respect <math>\angle A \implies \angle BAL = \angle CAM.</math> | ||
+ | |||
+ | <math>LE \perp AE, LF \perp AF \implies AELD</math> is cyclic. | ||
+ | |||
+ | <math>\angle CAM = \angle BAL = \angle FEL.</math> | ||
+ | |||
+ | <math>LE \perp AC \implies EF \perp AM.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Lemoine point line== | ||
+ | [[File:L M P line.png|430px|right]] | ||
+ | Let <math>\triangle ABC</math> be given. Let <math>L</math> be the Lemoine point of <math>\triangle ABC.</math> | ||
+ | |||
+ | Let <math>AH</math> be the height, <math>AM</math> be the median, <math>LD \perp BC, D \in BC,</math> | ||
+ | |||
+ | <math>LE \perp AC, E \in AC, LF \perp AB, F \in AB, P</math> be the midpoint <math>AH</math>. | ||
+ | |||
+ | Prove that the points <math>L, P,</math> and <math>M</math> are collinear. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>O</math> the circumcenter <math>\odot AELF, LO = AO.</math> | ||
+ | |||
+ | Denote <math>T</math> the midpoint <math>FE \implies OT \perp FE.</math> | ||
+ | |||
+ | <math>L</math> is centroid of <math>\triangle DEF \implies DLT</math> is <math>D-</math>median of <math>\triangle DEF.</math> | ||
+ | |||
+ | Denote <math>Q</math> the point symmetric <math>L</math> with respect <math>T \implies QT</math> is the midline of <math>\triangle LAQ \implies AQ \perp EF \implies Q \in AM \perp EF.</math> | ||
+ | |||
+ | <math>LD = 2 TL \implies DL = LQ \implies ML</math> is the median of <math>\triangle MDQ.</math> | ||
+ | |||
+ | <math>MP</math> is the median of <math>\triangle MHA, HA || DQ \implies</math> the points <math>L, P,</math> and <math>M</math> are collinear. | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 17:03, 31 July 2024
The reflecting of the median over the corresponding angle bisector is the symmedian. The angle formed by the symmedian and the angle bisector has the same measure as the angle between the median and the angle bisector, but it is on the other side of the angle bisector. The symmedian is isogonally conjugate to the median
There are three symmedians. They are meet at a triangle center called the Lemoine point.
Contents
Proportions
Let be given.
Let be the median,
Prove that iff is the symmedian than
Proof
1. Let be the symmedian. So Similarly
By applying the Law of Sines we get Similarly,
2.
As point moves along the fixed arc from to , the function monotonically increases from zero to infinity. This means that there is exactly one point at which the condition is satisfied. In this case, point lies on the symmedian.
Similarly for point
Corollary
Let be the symmedian of
Then is the symmedian of is the symmedian of is the symmedian of
vladimir.shelomovskii@gmail.com, vvsss
Symmedian and tangents
Let and it’s circumcircle be given.
Tangents to at points and intersect at point
Prove that is symmedian of
Proof
Denote WLOG, is symmedian of
Corollary
Let and it’s circumcircle be given.
Let tangent to at points intersect line at point
Let be the tangent to different from
Then is symmedian of
vladimir.shelomovskii@gmail.com, vvsss
Lemoine point properties
Let be given. Let be the Lemoine point of
Prove that is the centroid of
Proof
Let be the centroid of
The double area of is
Point is the isogonal conjugate of point with respect to
Similarly, one can get
The double area of is
Similarly, one can get is the centroid of
Corollary
Vector sum
Each of these vectors is obtained from the triangle side vectors by rotating by and multiplying by a constant
vladimir.shelomovskii@gmail.com, vvsss
Parallel lines
Let and it’s Lemoine point be given.
Let be an arbitrary point. Let be the foot from to line .
Denote the line through and parallel to
Denote the line parallel to such that distance and points and are both in the exterior (interior) of
Prove that points and are collinear.
Proof
Denote the foot from to .
Denote
Corollary
If squares and are constructed in the exterior of then where is the center of circle is the symmedian in through
vladimir.shelomovskii@gmail.com, vvsss
Radical axis
Circle passes through points and and touches line circle passes through points and and touches line Let be the Lemoine point of
Prove that the radical axis of these circles contains the symmedian of
Proof
Denote centers of and throught and respectively.
Denote line throught parallel to line throught parallel to
The ratio of distance from to to is equal to the ratio of distance from to to
is the orthocenter of
the radical axis of these circles contains the symmedian of
vladimir.shelomovskii@gmail.com, vvsss
Common Lemoine point
Let be given,
Let be the Lemoine point of
Prove that the point is the Lemoine point of
Proof
Denote point so that
Similarly denote and is the centroid of
(see Claim).
Let point be the centroid of is cyclic so therefore and are isogonals with respect
Similarly and are isogonals with respect
is the isogonal conjugate of a point with respect to a triangle
so is the Lemoine point of
Claim
Lines AP, BP and CP intersect the circumcircle of at points and
Points and are taken on the lines and so that (see diagram).
Prove that
Proof
is cyclic so
Similarly,
Similarly,
vladimir.shelomovskii@gmail.com, vvsss
Lemoine point extreme properties
Lemoine point minimizes the sum of the squares of the distances to the sides of the triangle (among all points internal to
Proof
Let us denote the desired point by Let us imagine that point is connected to springs of equal stiffness attached to the sides at points and and contacts sliding along them without friction. The segments modeling the springs will be perpendicular to the corresponding side. The energy of each spring is proportional to the square of its length. The minimum energy of the system corresponds to the minimum of the sum of the squares of the lengths of these segments, that is, the sum of the squares of the distances from to the sides.
It is known that the minimum spring energy corresponds to the equilibrium position. The condition of equilibrium at a point is the equality to zero of the vector sum of forces applied from the springs to the point The force developed by each spring is proportional to its length, that is, the equilibrium condition is that the sum of the vectors It is clear that the point corresponds to this condition.
vladimir.shelomovskii@gmail.com, vvsss
Lemoine point and perpendicularity
Let be given. Let be the Lemoine point of
is the midpoint
Prove that
Proof
is isogonal conjugated with respect
is cyclic.
vladimir.shelomovskii@gmail.com, vvsss
Lemoine point line
Let be given. Let be the Lemoine point of
Let be the height, be the median,
be the midpoint .
Prove that the points and are collinear.
Proof
Denote the circumcenter
Denote the midpoint
is centroid of is median of
Denote the point symmetric with respect is the midline of
is the median of
is the median of the points and are collinear.
vladimir.shelomovskii@gmail.com, vvsss